Question

A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. What is the specific heat capacity of iron?

Answer 1

`0.46"J"/("g" ""^@"C")`

Explanation:

Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.

A substance's specific heat tells you how much heat is needed in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.

`color(blue)("specific heat" = "heat"/("unit of mass " xx " unit of temperature"))`

Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.

Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.

• Scenario 1

For instance, let's assume that adding `"1086.75 J"` of heat to a `"15.75-g"` sample of iron will only increase its temperature by `1^@"C"`. In this case, the amount of heat needed per gram of iron will be

`"1086.75 J"/"15.75 g" = "69 J/g"`

In this scenario, adding `"69 J"` of heat for every gram of iron will increase its temperature by `1^@"C"`.

• Scenario 2

Now let's assume that this much heat would increase the temperature of `"1 g"` of iron by

`DeltaT = 175^@"C" - 25^@"C" = 150^@"C"`

In this case, the amount of heat needed per degree Celsius will be

`"1086.75 J"/(150^@"C") = "7.245 J/"""^@"C"`

In this scenario, you will get a `1^@"C"` increase in temperature by adding `"7.245 J"` for every `"1 g"` of iron.

• In real life

But since you know that adding that much heat will increase the temperature of `"15.75 g"` of iron by `150^@"C"`, you can say that

`"1086.75 J"/("15.75 g" * 150^@"C") = 0.46"J"/("g" ""^@"C")`

And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.

The equation that you'll be using from now on looks like this

`color(blue)(q = m * c * DeltaT)" "`, where

`q` - the amount of heat added to the sample
`m` - the mass of the sample
`c` - its specific heat
`DeltaT` - the change in temperature